The molecularity of a step is all the reactants that are present.
Unimolecular
A -> P (Rate =k[A])
Bimolecular
A+A -> P (Rate =k[A]^2)
A+B -> P (Rate =k[A][B])
Termolecular
A+A+A-> P (Rate =k[A]^3)
A+B+C-> P (Rate =k[A][B][C])
Hope that helps!
Search found 64 matches
- Sat Mar 16, 2019 10:06 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Elementary steps
- Replies: 3
- Views: 648
- Sat Mar 16, 2019 10:03 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: fast/slow step and reaction constant
- Replies: 1
- Views: 517
Re: fast/slow step and reaction constant
For multi-step processes, the faster steps will have a larger k while slower steps will have smaller one. However, these k values only represent that step and therefore, not the whole process. The overall k you calculate will be the same as the slow k if it is the first step, but if not it will be a...
- Sat Mar 16, 2019 9:59 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Test 2 Question 5
- Replies: 5
- Views: 886
Re: Test 2 Question 5
The correct answer is that the pH of 7 is basic at the new temperature. This is because the new neutral pH decreses to approx. 6.8. Since 7 is greater than that it will be basic. Hope that helps.
- Sun Mar 10, 2019 6:42 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Orders
- Replies: 2
- Views: 266
Re: Orders
You should pay attention to their molecularities, which is similar to adding the reactants bit not entirely the same.
For example, A+A -> P and A+B -> P are both Bimolecular but do not provide for the same rate law. Rate = k[A]^2 and Rate =k[A][B], respectively.
For example, A+A -> P and A+B -> P are both Bimolecular but do not provide for the same rate law. Rate = k[A]^2 and Rate =k[A][B], respectively.
- Sun Mar 10, 2019 6:34 pm
- Forum: Balancing Redox Reactions
- Topic: Question from textbook
- Replies: 3
- Views: 329
Re: Question from textbook
Since it states that it occurs in a basic medium you can also assume the presence of OH- molecules.
- Sun Mar 10, 2019 6:20 pm
- Forum: First Order Reactions
- Topic: 7th edition 7B.7
- Replies: 1
- Views: 262
Re: 7th edition 7B.7
You are given the reaction is a 1st order. For (a) and (b) you can simply multiply the half life by 2 then 3. For example, if we start off with 10M it takes 355 seconds to get to 5M or 1/2. It would take another 355 seconds to get to 2.5M or 1/4 and another 355 seconds to get to 1.25 or 1/8. For val...
- Sun Mar 03, 2019 6:04 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: K Value
- Replies: 7
- Views: 725
Re: K Value
I don't think this is true, it might have to do with the significant figures being used in the examples you have seen.
- Sun Mar 03, 2019 5:46 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: kinetic control
- Replies: 2
- Views: 377
Re: kinetic control
There is thermodynamic control and kinetic control. These deal with the composition of a reaction, in regards to products and reactants. Professor Lavelle went touched over this when we talked about C(Diamond) -> C(Graphite). This reaction is not thermodynamically favored, but is kinetically favored...
- Sun Mar 03, 2019 5:33 pm
- Forum: Second Order Reactions
- Topic: example in class
- Replies: 2
- Views: 258
Re: example in class
A second order reaction is one where the rate of reaction is proportional to the square of a reactant concentration. This is given by the equation: rate = k{A]^2 (Simplest form) The square of the concentration value signifies the interactions between molecules/atoms that define A(reactant). "k&...
- Sun Feb 24, 2019 7:22 pm
- Forum: Balancing Redox Reactions
- Topic: 1/2 Rxns: Elements vs. Compounds
- Replies: 3
- Views: 315
Re: 1/2 Rxns: Elements vs. Compounds
For writing half reactions you want to make sure that when you write it it serves its purpose. The half-reactions serve to show which element/compund is being oxidized or reduced within a reaction. Therefore, if an element within a compund is being oxidized/reduced you simply use the element. I have...
- Sun Feb 24, 2019 7:14 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Temperature's Effect on Gibbs Free Energy
- Replies: 2
- Views: 236
Re: Temperature's Effect on Gibbs Free Energy
There are some cases where the Gibbs free energy can be both + or - based on the value of T. >For example if both enthalpy and entropy are negative than at higher values of T, the reaction will be non-spontaneous. But, at lower temperatures it will be spontaneous. Hypothetical values deltaG = -300 -...
- Sun Feb 24, 2019 7:04 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6M.5 a)
- Replies: 1
- Views: 185
Re: 6M.5 a)
The left of the cell diagram represents the anode and the right is the cathode. Knowing this we can determine that Hg is being oxidized since the anode is where oxidization occurs and NO3 is being reduced since this is where reduction occurs. H+ atoms in the equation combine with a oxygen atoms from...
- Sun Feb 17, 2019 10:03 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: spontaneity
- Replies: 3
- Views: 342
Re: spontaneity
Yes, the only 2 forces are enthalpy and entropy, these values are true to the chemicals within the reaction, already. That is why the values of H(rxn) and S(rxn) are products minus reactants. External energy would be heat or work done on the system in order for the reaction to favor the production o...
- Sun Feb 17, 2019 9:50 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Elements with Gibbs Free Energy
- Replies: 1
- Views: 216
Re: Elements with Gibbs Free Energy
Elements in their standard state have an enthalpy of formation(deltaH) equal to 0. This does not mean Gibbs.
deltaH(formation) = 0
Then,
deltaG = deltaH -TdeltaS
deltaG = 0 - TdeltaS
Delta G is equal to 0 when the reaction is at equilibrium, which also means when K=1.
Hope this helps.
deltaH(formation) = 0
Then,
deltaG = deltaH -TdeltaS
deltaG = 0 - TdeltaS
Delta G is equal to 0 when the reaction is at equilibrium, which also means when K=1.
Hope this helps.
- Sun Feb 17, 2019 9:45 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: vibrational entropy
- Replies: 2
- Views: 283
Re: vibrational entropy
I'm sure there is a way to measure vibrational entopy, since I believe depends on the temperature of the system. Vibrational entropy studies have been done on some alloys (solids), but I am not too sure how accurate measurements are.
- Sun Feb 10, 2019 6:40 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Percent ionization vs percent deprotonation
- Replies: 3
- Views: 644
Re: Percent ionization vs percent deprotonation
These are the same in regards to Acid - Base equilibria. Deprotinization refers to acids losing a hydrogen proton to form H3O+ as a product, this cannot relate to bases since they do not lose protons. Acid: HCl + H2O -> H30+ +Cl- (deprotonization and ionization) Base: NaOH + H2O -> OH- + Na+ (unbala...
- Sun Feb 10, 2019 6:35 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: when to use kj and kj.mol^-1
- Replies: 2
- Views: 1777
Re: when to use kj and kj.mol^-1
It depends on the question, you can see them in both the kJ or kJ/mol format. You should know that the basic unit is Joules(J or kJ), as seen by the expression below where q is equal to enthalpy at constant pressure. ΔH=qp However, questions can word it in the following ways to signify kJ/mol. Molar...
- Sun Feb 10, 2019 6:23 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Problem #8.59 (Sixth Edition)
- Replies: 3
- Views: 374
Re: Problem #8.59 (Sixth Edition)
The standard reaction enthalpy of N2 is 0. This is because thisvreaction are spontaneous, 2N -> N2. This is also true for other diatomics like O2. Hope that helps.
- Sun Feb 03, 2019 5:14 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Predicting Spontaneity
- Replies: 4
- Views: 1415
Re: Predicting Spontaneity
Well you will want to remember the general rule of thumb of the reaction types, although, not always true. Exothermic reactions tend to be spontaneous since the change in ethalpy is negative and entropy increases. Vice versa for an endothermic reactions. This rule of thumb is helpful in checking you...
- Sun Feb 03, 2019 4:59 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Enthalpy defintions
- Replies: 2
- Views: 236
Re: Enthalpy defintions
The solution enthalpy is the enthalpy change when 1 mole of an ionic substance dissolves in water.
The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolve in water.
Hope that helps.
The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolve in water.
Hope that helps.
- Sun Feb 03, 2019 4:52 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Friday 2/1 Lecture Notes
- Replies: 2
- Views: 318
Re: Friday 2/1 Lecture Notes
I believe this was the combustion of glucose. I think it was used to demonstrate that combustion typically results in these products. Hope that helps.
- Sun Jan 27, 2019 5:57 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Method
- Replies: 2
- Views: 1459
Re: Hess's Method
Hess's Method does not deal with the breaking and forming of bonds. Hess's Law simply states that total enthalpy of a reaction is equal to the net change of enthalpy for the series of reactions/steps to arrive at the net equation. Typically, you will be given several reactions that will will work in...
- Sun Jan 27, 2019 5:37 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Capacity
- Replies: 8
- Views: 731
Re: Heat Capacity
Heat capacity is an extensive property. However, specific heat capacity is an intensive property, which you get when you divide heat capacity by mass.
Hope that helps!
Hope that helps!
- Sun Jan 27, 2019 5:32 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy Concept
- Replies: 1
- Views: 250
Re: Gibbs Free Energy Concept
You are right about Gibbs free energy being the energy available to be used. When energy is released it can change total energy of the surroundings and a release in energy would indicate an exothermic process.
- Sun Jan 20, 2019 7:34 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Changes in Temperature
- Replies: 8
- Views: 1094
Re: Changes in Temperature
Temperature is the only variable that can change the K value. If the temperature rises the products in an endothermic reaction will be favored and reactants in an exothermic reaction will be favored. Vice versa when the temperature is decreased. '
Hope this helps!
Hope this helps!
- Sun Jan 20, 2019 4:40 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 6th Edition 12.35
- Replies: 1
- Views: 294
Re: 6th Edition 12.35
K2 values are usually provided for polyprotic acids and this is seen by the HSeO4-. The minus indicates that this acid has lost a proton. This problem does not need to be done differently but with polyprotic acids, you will usually have to find more than one H3O+ concentration.
- Sun Jan 20, 2019 4:26 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition Problem 81
- Replies: 1
- Views: 79
Re: 6th Edition Problem 81
Although the second value is small, you still should find it. Some polyprotic acids have 3 de-protonizations, so It would be beneficial to calculate it for each one.
- Sun Jan 13, 2019 7:59 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Q and K
- Replies: 8
- Views: 574
Re: Q and K
K is an equilibrium constant at a certain temperature. This represents the ratio between [products] and [reactants] at equilibrium. Q represents the ratio between [products] and [reactants] at any point within the reaction. Therefore, Q=K at equilibrium since the two ratios will be equivalent. If Q ...
- Sun Jan 13, 2019 7:53 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Change in Pressure
- Replies: 4
- Views: 277
Re: Change in Pressure
You want to think of change in pressure as a change in volume. This will affect the reaction by changing concentrations and pushing the system away from equilibrium. Therefore, with the new concentration values, you will be solving for Q rather than K. By comparing the two you can determine which wa...
- Sun Jan 13, 2019 7:50 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Change in Pressure
- Replies: 1
- Views: 167
Re: Change in Pressure
K does not change with a change in pressure. There are two forms that pressure can increase, by increasing pressure through pumping in inert gasses, as Professor Lavelle mentioned, or by decreasing volume. However, changing pressure itself will not change concentration and therefore not affect the v...
- Sun Jan 13, 2019 5:22 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: converting from pressure to concentration
- Replies: 4
- Views: 419
Re: converting from pressure to concentration
R is constant but has many values. The 2 common values are R = 0.0821 L·atm/mol·K or R = 8.3145 J/mol·K. It is important to know the difference because the units are different. At the moment we are working with the first one where atm = atmospheres.
Hope that helps!
Hope that helps!
- Sat Dec 08, 2018 7:13 pm
- Forum: Properties of Light
- Topic: Hertz
- Replies: 1
- Views: 445
Re: Hertz
You would want to use cycles per second because you will use that to cancel with the constants to give you the right units for your answer.
- Sat Dec 08, 2018 7:05 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: the mole
- Replies: 4
- Views: 548
Re: the mole
The equation for density is density = mass/volume. I believe that your explanation is correct.
- Sat Dec 08, 2018 7:02 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Ka
- Replies: 1
- Views: 206
Re: Ka
I don't think we will be using Ka values in 14a but we will be in 14b. It probably won't be on the final because no Ka values will be provided to compare acids and I believe Professor Lavelle mentioned we won't be calculating it. I would recommend knowing the concept behind it and knowing pKa and pKb.
- Sun Dec 02, 2018 5:31 pm
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: LDF
- Replies: 7
- Views: 1663
Re: LDF
LDFs are found in every molecule since the only requirement to create it is chance - electrons being concentrated on one side of an atom/molecule.
Hope that helps!
Hope that helps!
- Sun Dec 02, 2018 5:30 pm
- Forum: Bronsted Acids & Bases
- Topic: Bronsted and Lewis Acids and Bases
- Replies: 3
- Views: 197
Re: Bronsted and Lewis Acids and Bases
The Lewis definition deals with electrons while the Bronsted-Lowry deals with protons. A Lewis acid is a lone pair acceptor while a Lewis base is a lone pair donor, lone pair indicating electrons. A B-L acid is a proton donor while a B-L base is a proton acceptor, proton being an H+ ion. Hope that h...
- Sun Dec 02, 2018 4:23 pm
- Forum: Hybridization
- Topic: Hybridization Question
- Replies: 1
- Views: 128
Re: Hybridization Question
There is a debate about hybridization involving d-orbitals since its energy level is far removed from that of s or p but the hybridization involving 7 regions would be sp3d3. I don't think we need to know that in the slightest but the molecule IF7 has a pentagonal bipyramidal structure with a hybrid...
- Sun Nov 25, 2018 11:04 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Types of Lone Pairs
- Replies: 2
- Views: 310
Re: Types of Lone Pairs
These lone pair names relate to where they are on the structure. An axial lone pair will be on one end of the axis while an equatorial lone pair will not be on the axis. The axis is the combination of the 2 surrounding atoms and the central atom, with a bond angle of 180 degrees similar to a x-axis ...
- Sun Nov 25, 2018 3:05 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: Molecular Orbital Theory
- Replies: 1
- Views: 181
Re: Molecular Orbital Theory
Molecules without resonance are known to have localized electrons, which means they stay between two atoms. You can think of it as a molecule with different atoms bonded to a central atom. The atom with the highest electronegativity will want to keep the electrons, whereas in a resonance structure y...
- Sun Nov 25, 2018 2:58 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Chapter 4, Question 7 (6th Edition)
- Replies: 1
- Views: 130
Re: Chapter 4, Question 7 (6th Edition)
The OSCL relates to bond angles and not the structure, in a trigonal pyramidal structure, these will be the same. A trigonal pyramidal can only form when there are 4 areas of electron density and that would indicate a tetrahedral structure. As you said, there are only 3 surrounding atoms, but the 4t...
- Sun Nov 18, 2018 6:28 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Writing Bond Angles with Multiple Central Angles
- Replies: 2
- Views: 253
Re: Writing Bond Angles with Multiple Central Angles
Great question! I believe Dr. Lavelle had mentioned this and it is a rather simple concept, at least in your example, since to know what bond angle to look at you need 3 atoms. C-C wouldn't be of any interest leaving only H-C-C. I think this can be explained with a mathematical concept, where two po...
- Sun Nov 18, 2018 6:11 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular Geometry
- Replies: 6
- Views: 596
Re: Molecular Geometry
You can think of molecular geometry as all the bonded regions in a molecule. Then you can think of electron density geometry as all the bonded regions and the lone pairs in the molecule. These will be the same if there are no lone pairs in the molecule.
- Sun Nov 18, 2018 5:25 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.7 6th edition
- Replies: 2
- Views: 2673
Re: 4.7 6th edition
The first step you want to take is drawing the Lewis structure of the molecule, using the element with the lowest ionization in the middle you can build a structure surrounding sulfur - as you mentioned. Since the molecule has no charge, the optimal scenario is having all formal charges equalling 0....
- Sun Nov 11, 2018 9:34 pm
- Forum: Bond Lengths & Energies
- Topic: Hydrogen Bonds
- Replies: 1
- Views: 277
Re: Hydrogen Bonds
This is caused by the difference of electronegativities between Nitrogen and Fluorine. Since fluorine has the highest electronegativity of any element it is very unwilling to lose electrons. Therefore, the partial charge associated with fluorine greater than the one between nitrogen, both being nega...
- Sun Nov 11, 2018 4:48 pm
- Forum: Resonance Structures
- Topic: Calculating Bond Length
- Replies: 3
- Views: 792
Re: Calculating Bond Length
It would be a guess between the values since we would need instruments to measure the hybrid structure formed by the resonance structures. For example, if a molecule has twos single bonds and 1 double bond it with a single bond length of 100 and a double bond length of 80, you would simply say 90 fo...
- Sun Nov 11, 2018 4:20 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 12
- Views: 1099
Re: Bond Angles
I'm am pretty sure we will. Bond angles can be determined after finding the molecular structure of an atom. Certain structures have certain angles. For example, a tetrahedral structure will always have bond angles of ~109.5 degrees and a square planar structure will always have bind angles of 90 deg...
- Sun Nov 11, 2018 4:14 pm
- Forum: Lewis Structures
- Topic: Hypervalent compound vs variable covalence
- Replies: 2
- Views: 361
Re: Hypervalent compound vs variable covalence
For the most part, the terms are very similar. The difference that comes to mind is that a hypervalent compound is referring to a chemical structure that defies the octet rule, such as PCL5. Whereas variable covalency is a characteristic that can be associated with atoms starting from the 3p orbital...
- Sun Nov 04, 2018 4:45 pm
- Forum: Lewis Structures
- Topic: Hydrogen Sulfide
- Replies: 1
- Views: 110
Re: Hydrogen Sulfide
HSO3- actually has two structures you can form, however, one does not display acidic properties while one does. When H is bonded to O the molecule is acidic and created hydrogen sulfide. When H is bonded to S it creates "sulfonate", which is not acidic. The thing to remember when working w...
- Sun Nov 04, 2018 4:33 pm
- Forum: Lewis Structures
- Topic: 7th edition 2B.9
- Replies: 4
- Views: 387
Re: 7th edition 2B.9
You draw them separately because they either are ionic or have a polyatomic ion. NH4+ is ammonium, which is polyatomic. K3P is ionic so electrons are being "transferred". CLO- is hypochlorite, which is also polyatomic.
- Sun Nov 04, 2018 4:26 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: More exceptions for electron configurations??
- Replies: 1
- Views: 266
Re: More exceptions for electron configurations??
Yes, the other elements in the Chromium and Copper column also exhibit this exception, however, those two are the only ones we need to know for this class.
- Sun Nov 04, 2018 4:23 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge SO42-
- Replies: 3
- Views: 11509
Re: Formal Charge SO42-
That's a great question. You want to make the most '0' formal charges to create the most stable structure in a molecule. However, the sum of formal charges should always equal the charge of the molecule. Therefore, adding double bonds to the other oxygen atoms will transfer make the FC on Sulfur -2,...
- Sun Oct 28, 2018 6:39 pm
- Forum: Ionic & Covalent Bonds
- Topic: Polyatomic Ions [ENDORSED]
- Replies: 4
- Views: 501
Re: Polyatomic Ions [ENDORSED]
You will most likely be better off knowing the common ones. If they aren't provided it will hugely benefit you and if they are it will save a lot of time just knowing them.
- Sun Oct 28, 2018 6:38 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Noble Gases
- Replies: 3
- Views: 330
Re: Noble Gases
All you would do is use the last noble gas as your place-holder. So for Argon, the shorthand would be [Ne]3s2 3p6
Hope this helps!
Hope this helps!
- Sun Oct 28, 2018 6:23 pm
- Forum: Ionic & Covalent Bonds
- Topic: the Octet rule
- Replies: 21
- Views: 4009
Re: the Octet rule
Most atoms do abide by the octet rule, however, there are exceptions. Hydrogen and Helium are examples of atoms that do not since they do not require are fine with having only 2 electrons. There are also examples of atoms having more than 8 electrons, such as sulfur in SF6, this is known as an expan...
- Sun Oct 21, 2018 3:18 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Nodes
- Replies: 1
- Views: 199
Re: Nodes
The atomic spectra provides evidence that nodes or areas of 0 electron density exist. This is because the emissions support a quantized energy level model. Think of a node as the space between two orbits drawn around a nucleus, and since an electron cannot go from n=5 to n'=3.4 there will be no emis...
- Sun Oct 21, 2018 3:05 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Shielding
- Replies: 5
- Views: 1127
Re: Shielding
Shielding within multi-electron atoms results from several factors. In hydrogen the presence of only one electron results in the electron feeling the full nuclear charge. However, in larger molecules electrons in orbitals further from the nucleus feel an effective nuclear charge. This is caused by r...
- Sun Oct 21, 2018 2:57 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Differentiating spectroscopies
- Replies: 2
- Views: 173
Re: Differentiating spectroscopies
Atomic spectroscopy concerns an element and is focused on the natures of only that element, such as exciting an electron. Molecular spectroscopy concerns molecules, which are composed of atoms. Molecular spectroscopy focuses on how molecules interact including their bonding and electromagnetic inter...
- Sun Oct 14, 2018 11:38 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric effect and atomic spectra difference
- Replies: 3
- Views: 318
Re: Photoelectric effect and atomic spectra difference
The atomic spectra isn't completely related to light shining on a metal. In the photoelectric effect scientists were looking to eject electrons, while the atomic spectra the electron is still in the vicinity of the nucleus but at a higher energy level. These electrons will return to the ground state...
- Tue Oct 09, 2018 7:55 pm
- Forum: Properties of Light
- Topic: 1.A.11 Atomic Spectra
- Replies: 1
- Views: 109
Re: 1.A.11 Atomic Spectra
The 3 main series of the hydrogen spectrum are Lyman, Balmer and Paschen series. Each series are contains several 'lines' in the spectrum and you are right that they are related to the same energy level. Lyman being n'=1, Balmer being n'=2, and Paschen being n'=3. Also each grouping represents diffe...
- Tue Oct 09, 2018 7:40 pm
- Forum: Properties of Electrons
- Topic: Behavior of Electrons and Energy Levels
- Replies: 3
- Views: 194
Re: Behavior of Electrons and Energy Levels
Thats a really interesting question! The electron jumping from one level to the next is a simplification of the process and I think you are thinking of it in the Bohr Model. If you were to look at it in the VESPR or s, p, d, f model the path an electron would take to an excited state is hard to thin...
- Tue Oct 09, 2018 7:31 pm
- Forum: Properties of Light
- Topic: Diffraction patterns
- Replies: 2
- Views: 121
Re: Diffraction patterns
For diffraction patterns to arise there must be a diffraction object, this object like a slit causes a shift in the 'waves' of light and this shift leads to waves interfering with one another. The patterns are caused by countless number of interactions which graphics can't really depict. To answer y...
- Wed Oct 03, 2018 11:11 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: G.25 Molarity of Solution after Diluting
- Replies: 1
- Views: 103
Re: G.25 Molarity of Solution after Diluting
A useful equation you could use is the exponential function and in this case - (10)(2)^90, which should give you a value of 1.23794 x 10^28 ml or 1.23794 x 10^25 L. Once you find the number of moles in the initial concentration you can use the Molarity equation to find your answer because the number...
- Wed Oct 03, 2018 10:55 pm
- Forum: Limiting Reactant Calculations
- Topic: Limiting Reactant Steps General Question
- Replies: 3
- Views: 374
Re: Limiting Reactant Steps General Question
Great question! We use an "imaginary" value of 100g to simplify calculations that need to be made to isolate the amount of moles of a certain element when we are dealing with mass percentages. Since percentages are proportional it does not matter what the mass we use is. This is also true ...
- Wed Oct 03, 2018 10:41 pm
- Forum: Significant Figures
- Topic: Sig Figs when finding Mol
- Replies: 4
- Views: 9096
Re: Sig Figs when finding Mol
A good rule to remember when dealing with significant figures, especially in division and multiplication problems, is that your answer cannot be more significant than the least significant value present in the equation. For example, if you divide 4.56 by 2.1 (i.e. 4.56/2.1), then your answer would b...